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数学公式

\begin{equation} x=\sqrt{b} \end{equation}

If $a^2=b$ and ( b=2 ), then the solution must be either $$ a=+\sqrt{2} $$ or $$ a=-\sqrt{2} $$.

$$J(\theta) = \frac{1}{2m}\sum_{i=1}^{m}(\theta^{T}X_{i} - Y_{i})^2$$

$e=mc^2$

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}

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@end

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